Question

Find two consecutive positive integers,sum of whose squares is 365

Answer 1

Heya user, 

Consider the two consecutive integers as a,(a+1)
Then,
----------> a² + (a+1)² = 365
=====> 2a² + 2a + 1 = 365
=====> a² + a = 182
=====> a² + a - 182 = 0
=====> a² + 14a - 13a - 182 = 0
=====> ( a + 14 )( a - 13 ) = 0
=====> a = -14, 13;

Since, a is a positive integer, a = 13;
Hence the two consecutive integers are 13, 13+1 = ( 13 , 14 ) <-- Ans.

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assume the number be m and m + 1

★Situation :-

m² + (m + 1)² = 365

m² + m² + 2m + 1 = 365

2m² + 2m - 364 = 0

${\boxed{\sf\:{Take\;2\;as\;common\;in\;LHS\;we\; get}}}$

m² + m + 182 = 0

★Factorise it we get :-

m² + 14m - 13m - 182 = 0

m(m + 14) - 13(m + 14) = 0

(m + 14)(m - 13) = 0

m + 14 = 0

m = -14

m - 13 = 0

m = 13

$\text{Here\;m\;is\;an\;positive\;integer\;so\;it\;can't\;be\; negative}$

★Hence ,

${\boxed{\sf\:{Two\;numbers\;are\;13\;and\;14}}}$

$\boxed{\begin{minipage}{14 cm} Additional Information \\ \\ \ A\; Quadratic\; Equation\;has\;three\;equal\;roots \\ \\ 1)Real\;and\;Distinct \\ \\ 2)Real\;and\;Coincident \\ \\ 3) Imaginary \\ \\ Note:-Third\; Imaginary\;is\;not\;taken\;in\;class\;10th \\ \\ If\;p(x)\;is\;a\; quadratic\; polynomial\;then\;p(x)=0\;is\;called\; Quadratic\; Polynomial \\ \\ General\;Formula=ax^2+bx+c=0 \\ \\ A polynomial\;whose\;degree\;will\;be\;2\;is\; considered\;as\; Quadratic\; Polynomial \\ \\ Rules\;for\; solving\; Quadratic\; Equations:- \\ \\ Put\;all\;the\;terms\;into\;RHS\;and\;make\it\;zero \\ \\ Substitute\;all\; factors\;equal\;to\;Zero\;Get\;a\;equal\; solution \end{minipage}}$